Transcript Document

Molecular Orbital Theory
Diatomic molecules: Heteronuclear molecules
Chem 59-250
In heteronuclear diatomic molecules, the relative contribution of atomic orbitals to
each MO is not equal. Some MO’s will have more contribution from AO’s on one
atom than from AO’s on the other. This means that the coefficients in the MO will
not be the same! For example, in hydrogen fluoride (HF), some orbitals are derived
more from H than F and vice versa. The more the contribution from AO’s on a given
atom, the higher the coefficient in front of the AO in the MO.
The combinations of  symmetry: (note that the 1s orbital on H is closer in energy to the
2pz orbital on F so we will look at that combination because there will be more interaction)
This MO is more F-like
+
1sH
2pzF
2 = ( 0.1 1sH +  0.9 2pzF)
+
1sH
2pzF
This MO is more H-like
3* = ( 0.9 1sH -  0.1 2pzF)
Chem 59-250
Molecular Orbital Theory
Because the contributions are not equal, the MO diagram will be skewed.
MO diagram for HF
There is a little bit of mixing
between the H 1s and the F 2s
orbital but it interacts mostly with
the 2pz.
F also has the 2px and 2py
orbitals that cannot interact with
the H 1s orbital because they
have the wrong symmetry! If you
try to combine these orbitals with
the 1s on H, you will find that the
overlap integral, S, is equal to 0.
Thus these orbitals are
exclusively found on the F atom
and are called non-bonding. The
energies of these orbitals do not
change from the energies in the
F atom.
The orbitals that are derived mostly from F are going to be closer to the energies of
the atomic orbitals of F and vice versa.
Chem 59-250
Molecular Orbital Theory
MO diagrams for other heteronuclear diatomics are formed in exactly the same way
as that of H-F or those of the homonuclear diatomics: atomic orbitals of appropriate
symmetry will interact to produce MO’s. The orbtals that are closest in energy to one
another will interact the most - i.e. there will be greater stabilization of the bonding
MO and destabilization of the antibonding MO.
In this diagram, the labels are
for the valence shell only - it
ignores the 1s shell. Note that
the 1p and 3 MO’s are not
labeled correctly. Also notice,
there is no g or u because the
molecule does not have a
center of symmetry.
Again, the MO’s will have a larger
contribution from one of the atoms.
In CO, the HOMO has more of a
contribution from C AO’s, so CO
acts as an electron donor through
the carbon atom.
The relative energies of the
AO’s can be estimated based
on the electronegativity of the
atoms. More electronegative
atoms will have more stable
(lower) orbitals.
Molecular Orbital Theory
Polyatomic molecules: The bonding in H3+
Each H atom has only a 1s orbital, so to obtain MO’s for the H3+ cation, we
must make linear combinations of the three 1s orbitals.
Chem 59-250
Consider the situation where the H atoms arranged to make a linear geometry. Note
that atoms that are related by symmetry must be treated together.
+
1sA
1sB
+
1sA
-
no
contribution
1sC
-
1sB
1sC
1sB
1g = ( 0.25 1sA +  0.5 1sB + 0.25 1sC)
The 1s orbital of HB does not
have appropriate symmetry to
interact with the combination of
1sA - 1sC. 1 node so less stable
than 1g.
1u = ( 0.5 1sA -  0.5 1sC)
2 nodes so this is the least
stable of these MO’s.
-
+
1sA
0 nodes so most stable.
+
1sC
2g* = (- 0.25 1sA +  0.5 1sB - 0.25 1sC)
Molecular Orbital Theory
Polyatomic molecules: The bonding in H3+
For the linear cation, the MO diagram would then be:
Chem 59-250
Note that the three H atoms are held
together by a total of only two electrons.
Hcentral
H3+
H2(SALC)
H2
Since the terminal H atoms are symmetry
related and must be considered as a pair, we
must make symmetry adapted linear
combinations (SALC’s) of their orbitals to
interact with the central atom:
2g*
Energy
Opposite phases (u):
u
1u
g
1s
Same phases (g):
1s
This is the approach that we must use
for all polyatomic molecules.
1g
Note that the u combination will be a little bit
higher in energy than the g so the nonbonding LUMO is a bit higher in energy than
the AO in a free H atom.
BUT…since there are three atoms, a linear arrangement is not the only possibility!
Molecular Orbital Theory
Polyatomic molecules: The bonding in H3+
Chem 59-250
Consider the situation where the H atoms arranged in an equilateral triangular
geometry. Note that now all the atoms are related by symmetry must be treated
together as a set. The symmetry of this arrangement is D3h, which tells us the
symmetry of each of the orbitals and also that there must a pair of degenerate MO’s.
1sC
0 nodes so this is the most stable MO.
a1’ = (1/ 3 1sA + 1/ 3 1sB + 1/ 3 1sC)
+/- +/1sA
+/-
1sB
The doubly-degenerate pair of MO’s have e’
symmetry. Although they do not look alike, each
orbital in the pair has the same energy. Notice that
the orbitals have roughly the same symmetry as
would px and py orbitals in the middle of the H3 ring.
Each of these MO’s has one node so
the e’ orbital pair will be higher in
energy than the a1’ orbital.
e’ = (1/ 2 1sB - 1/ 2 1sC)
e’ = (2/ 6 1sA - 1/ 6 1sB - 1/ 6 1sC)
Molecular Orbital Theory
Polyatomic molecules: The bonding in H3+
For the triangular cation, the MO diagram would then be:
Chem 59-250
Energy
3H
Notice that the e’ pair must have identical energies
and that the pair is less stable than either the free
atoms or the a’ MO so it can be called anti-bonding.
H3+
1e’*
3  1s
Again, the three H atoms are held together by a
total of only two electrons.
Furthermore, there is no central atom and each H is
related by symmetry. Because of this, the use of
SALC’s and MO theory can provide us with a much
better model of the bonding than we could get from
VBT and the localized model of bonding.
1a’
BUT…in theory, we could have any arrangement in between the linear and the
triangular, so how do we find out which geometry is the most stable?
HA
HB
HC
HA
HB
HC
HB
HB
HA
HC
H
A
HC
Chem 59-250
Molecular Orbital Theory
Polyatomic molecules: The bonding in H3+
We can use a Walsh diagram to compare and assess the relative energies of
different possible structures. In a Walsh diagram, the relative energies of important
MO’s are plotted as the value of a metrical parameter (e.g. bond lengths or angles)
is changed. The amount of stabilization or destabilization of the MO’s is based on
the amount of increase or decrease in the in-phase overlap of the AO’s used to
make each molecular orbital.
Walsh diagram for Dh to D3h
In the example of H3+ there are only two electrons, thus we only have to examine how the energy
of the lowest orbital (1g to a1’) varies with the change in angle. From the diagram we can see that
the energy decreases from Dh to D3h so the electronic energy of the triangular arrangement will
be lower than that of the linear arrangement. This means that the triangular form will be the most
stable arrangement. If there were more electrons, we would have to consider how the energies
change for any orbital that might be populated with electrons.
Chem 59-250
Molecular Orbital Theory
Polyatomic molecules
The steps you can use to build a MO diagram for any
polyatomic molecule are:
1. Determine the symmetry of the molecule and figure
out which atoms are symmetry related.
2. Make appropriate symmetry adapted linear
combinations of atomic orbitals for the symmetry related
atoms.
3. Estimate the energies of the AO’s using the
electronegativities of the atoms.
4. Use symmetry to determine which orbitals can
interact with each other to form bonding and antibonding MO’s. Those that can’t interact will be nonbonding.
5. Arrange MO’s in order of increasing energy based on
the number of nodes and use the available electrons to
fill the lowest energy orbitals.
For simple molecules, you can draw pictures of the
SALC’s and MO’s (as shown) in the example. For more
complicated molecules, it is easier to just use symmetry
and character tables.
e.g. BeH2
Chem 59-250
Molecular Orbital Theory
Polyatomic molecules
If you are trying to estimate the appropriate geometry for a triatomic molecule using a Walsh
diagram, all that is necessary is to correctly determine the number of electrons that will populate
the orbitals in the diagram. Then you can estimate which electron configuration will provide the
lowest overall energy (and thus the most stable geometry).
E.g.’s:
There are 4 valence electrons in
BeH2 so these electrons can fill the
1g and the 1u orbitals in the
diagram (shown in blue). The lowest
energy combination is obtained
when the bond angle is 180°.
In H2O, there are 8 valence
electrons which will fill up all four
orbitals in the diagram. Since the
energy of the 1b1 orbital doesn’t
change as the angle is changed, the
overall energy is mostly determined
by the relative energies of the 2a1
and the 1b2. A reasonable guess is
shown in red.
Orbital overlap analyses such as these allow for the prediction of molecular geometry using the
delocalized model for covalent bonding in the same way that VSEPR is used in the localized
approach.
Chem 59-250
Molecular Orbital Theory
Polyatomic molecules
E.g. Buiding a MO diagram for NH3:
1. The point group is C3v.
2. All H’s are related so these must be split into SALC’s of a1
and e symmetry (these have exactly the same shape as
those in the example for H3+, but the point group that must
be used for NH3 is C3v).
3. XN  3 and XH  2.2 so the energy levels of the AO’s on N
will be lower than those on the H atoms.
4. From the C3v character table, the symmetry of the AO’s
on N are: A1(2s), A1(2pz), and E(2px,2py). Each of these
orbitals can interact with the SALC’s from the H atoms.
5. Fill the MO’s with the 8 valence electrons.
In NH3, the HOMO is a mostly nitrogen-based orbital that
corresponds to the lone pair of electrons from VBT. This is
why ammonia acts as a Lewis base at the N atom. The
LUMO is the 2e level that has more H character - this shows
why NH3 can also act as a Lewis acid through the H atoms.
HOMO
LUMO
Chem 59-250
Molecular Orbital Theory
Polyatomic molecules
E.g. Buiding a MO diagram for SiH4:
Si
1. The point group is Td.
2. All H’s are related so these must be split into SALC’s of a1
and t2 symmetry (these are harder to draw, so it’s easier to
use a symmetry analysis of the four H atoms to get the
SALC’s for AO’s of the H atoms).
SiH4
4H
2a1*
2t2*
3p
3. XSi  1.9 and XH  2.2 so the energy levels of the AO’s on
the H atoms will be lower than those on the Si atom.
4. From the Td character table, the symmetry of the AO’s on
Si are: A1(3s), T2(3px,3py,3pz). Each of these orbitals can
interact with the SALC’s from the H atoms.
5. Fill the MO’s with the 8 valence electrons.
In SiH4, the HOMO is a mostly based on the peripheral
hydrogen atoms and the LUMO is dominated by Si so SiH4
will act as an electron donor through the H atoms and an
electron acceptor at Si.
T2
1t2
3s
A1
1a1
Molecular Orbital Theory
Polyatomic molecules
Chem 59-250
Consider what happens to the -bonding MO’s if the symmetry is reduced from Td to C2v: there
will no longer be triply-degenerate MO’s. The four bonding orbitals are split into two sets - those
with more H character (higher in energy) and those with more Cl character (lower in energy).
C
CH4
4H
C
CH2Cl2
2 H, 2 Cl
4a1*
2a1*
2b2*
2t2*
2b1*
2p
3a1*
T2
2p
2H
1t2
A1
2s
2s
1b2
A1,B2
2a1
1a1
Since these diagrams are only concerned with the -bonding,
the AO’s and MO’s from the 3p orbitals on Cl have been omitted.
The correlation lines from the 2p orbitals are not drawn for
clarity.
2 Cl
A1,B1
1b1
1a1
Molecular Orbital Theory
Pi-bonding in polyatomic molecules
Chem 59-250
Molecular orbital theory considers p-bonding in exactly the same way as -bonding. Orbitals with
appropriate symmetry will interact to form bonding and anti-bonding MO’s. In contrast to Lewis
theory or VBT, resonance structures are not needed to describe the p-bonding because the MO’s
are spread equally over the p-system of the entire molecule. Not surprisingly, the delocalized
model of bonding provides a much better picture of the delocalized p-system.
E.g. the p-bonding in CO3-2
In the D3h molecule, the 2pz orbitals on the O
atoms give SALC’s of the following form:
These are analogous to the SALC’s we built
for H3+ and so the normalization coefficients
are identical:
a2” = (1/ 3 2pzA + 1/ 3 2pzB + 1/ 3 2pzC)
e” = (1/ 2 2pzB - 1/ 2 2pzC)
e” = (2/ 6 2pzA - 1/ 6 2pzB - 1/ 6 2pzC)
The 2pz orbital on C can only interact with the a2”
SALC to give bonding and anti-bonding MO’s.
A MO diagram of
the p-bonding:
Molecular Orbital Theory
Pi-bonding in aromatic molecules
Chem 59-250
The p-bonding in aromatic molecules is readily predicted using MO theory. Although the details of
the treatment that predicts the energies is better left to a higher level course, the symmetry and
relative energies of the MO’s are easily understood from the number of nodes in each of the
MO’s.
The p-bonding in C5H5-1 (aromatic)
The p-bonding in C3H3+1 (aromatic)
2 nodes
1 node
1 node
0 nodes
0 nodes
The p-bonding in C4H4+2 (aromatic)
2 nodes
The p-bonding in C6H6 (aromatic)
3 nodes
1 node
2 nodes
0 nodes
1 node
Aromatic compounds must have a completely filled set of bonding p-MO’s.
This is the origin of the Hückel (4N+2) p-electron definition of aromaticity.
0 nodes
Chem 59-250
Molecular Orbital Theory
“Infinite” polyatomic molecules
An adaptation of MO theory is also used to treat extended (essentially infinite) “molecular”
systems such as pieces of a metal (or an alloy) or a molecule such as a diamond. Remember
that whenever atomic orbitals have appropriate symmetry and energy, they will interact to form
bonding and anti-bonding MO’s. We have only considered small molecules but there are
essentially no limits to the number of atoms that can mix their corresponding AO’s together. This
mixing must occur because of the Pauli exclusion principle (i.e. so that no two electrons in the
solid have the same quantum numbers).
MO
anti-bonding
AO(1)
AO(2)
As the number of
interacting atoms (N)
increases, the new
bonding or anti-bonding
MO’s get closer in
energy.
bonding
In the extreme, the MO’s of a given type become so
close in energy that they approximate a continuum of
energy levels. Such sets of MO’s are known as
bands. In any given band, the most stable MO’s
have the most amount of bonding character and the
highest energy MO’s are the most anti-bonding. The
energy difference between adjacent bands is called
the band gap.
Chem 59-250
Molecular Orbital Theory
Remember that the closer to AO’s of appropriate symmetry are in energy, the more they interact
with one another and the more stable the bonding MO that will be formed. This means that as the
difference in electronegativity between two atoms increases, the stabilization provided by covalent
bonding decreases (and the polarity of the bond increases). If the difference in energy of the
orbitals is sufficiently large, then covalent bonding will not stabilize the interaction of the atoms. In
that situation, the less electronegative atom will lose an electron to the more electronegative atom
and two ions will be formed.
AO(1)
AO(1)
AO(1)
AO(1)
AO(2)
AO(2)
AO(2)
AO(2)
Most covalent
DX < 0.5 : covalent
Polar Covalent
2 > DX > 0.5 : polar
Ionic
DX > 2 : ionic